What is the highest oxidation state exhibited by group 17 elements?
A. $ + 1$
B. $ + 3$
C. $ + 5$
D. $ + 7$
Answer
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Hint: The oxidation state which is also known as oxidation number describes the degree of oxidation i.e., loss of electrons of an atom in a chemical compound. Conceptually, the oxidation state is a hypothetical charge that an atom would have if all the bonds of an atom of different elements are ionic and with no covalent character. It may have a value which can be positive, negative or zero.
Complete answer:
Group 17 also known as the halogen family consists of highly reactive nonmetals and includes fluorine, chlorine, bromine, iodine and astatine. The general electronic configuration of group 17 elements is $n{s^2}n{p^5}$ i.e., they consist of seven electrons in their outermost shell and are short of one electron to achieve the nearest noble gas configuration. Hence, all the halogens generally exhibit $ - 1$ oxidation state.
However, the elements such as chlorine, bromine and iodine also exhibit $ + 1$, $ + 3$, $ + 5$ and $ + 7$ oxidation states in various compounds. This higher oxidation state of chlorine, bromine and iodine is shown when these elements are bonded with high electronegative elements like fluorine and oxygen. The halogens in oxides and oxoacids of chlorine and bromine consist of $ + 4$ and $ + 6$ states.
Examples of halogens in compounds with higher oxidation states are as follows:
1. $ClF$: The oxidation state of chlorine is $ + 1$ in the given compound.
2. $Cl{F_3}$: The oxidation state of chlorine is $ + 3$ in the given compound.
3. $Br{F_5}$: The oxidation state of bromine is $ + 5$ in the given compound.
4. $I{F_7}$: The oxidation state of iodine is $ + 7$ in the given compound.
Thus, we can conclude that the highest oxidation state exhibited by group 17 elements is $ + 7$. Hence, option (D) is the correct answer.
Note:
It is important to note that fluorine is the most electronegative element and relatively smaller in size as compared to other halogens. So, it cannot exhibit any positive oxidation state. Unlike other halogen atoms, it does not have d-orbitals in its valence shell and hence, is unable to show higher oxidation states.
Complete answer:
Group 17 also known as the halogen family consists of highly reactive nonmetals and includes fluorine, chlorine, bromine, iodine and astatine. The general electronic configuration of group 17 elements is $n{s^2}n{p^5}$ i.e., they consist of seven electrons in their outermost shell and are short of one electron to achieve the nearest noble gas configuration. Hence, all the halogens generally exhibit $ - 1$ oxidation state.
However, the elements such as chlorine, bromine and iodine also exhibit $ + 1$, $ + 3$, $ + 5$ and $ + 7$ oxidation states in various compounds. This higher oxidation state of chlorine, bromine and iodine is shown when these elements are bonded with high electronegative elements like fluorine and oxygen. The halogens in oxides and oxoacids of chlorine and bromine consist of $ + 4$ and $ + 6$ states.
Examples of halogens in compounds with higher oxidation states are as follows:
1. $ClF$: The oxidation state of chlorine is $ + 1$ in the given compound.
2. $Cl{F_3}$: The oxidation state of chlorine is $ + 3$ in the given compound.
3. $Br{F_5}$: The oxidation state of bromine is $ + 5$ in the given compound.
4. $I{F_7}$: The oxidation state of iodine is $ + 7$ in the given compound.
Thus, we can conclude that the highest oxidation state exhibited by group 17 elements is $ + 7$. Hence, option (D) is the correct answer.
Note:
It is important to note that fluorine is the most electronegative element and relatively smaller in size as compared to other halogens. So, it cannot exhibit any positive oxidation state. Unlike other halogen atoms, it does not have d-orbitals in its valence shell and hence, is unable to show higher oxidation states.
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